# A slipping bowling ball

#### kingsmaug

1. Homework Statement

A spherical bowling ball with mass m = 4.3 kg and radius R = 0.102 m is thrown down the lane with an initial speed of v = 8.2 m/s. The coefficient of kinetic friction between the sliding ball and the ground is μ = 0.29. Once the ball begins to roll without slipping it moves with a constant velocity down the lane.

magnitude of angular acceleration during sliding: 69.66 rad/s^2
magnitude of linear acceleration during sliding: 2.84 m/s^2
how long to roll without slipping: 0.824 s
length of slide before rolling: 5.79 m
and after it begins to roll without slipping, the rotational kinetic energy is less than the translational kinetic energy.

What is the magnitude of the final velocity?

2. Homework Equations
KE=1/2mv^2
Erot=1/2Iw^2
w=v/r

3. The Attempt at a Solution
I assumed that the KE of the ball immediately before hitting would be equal to the sum of the final KE and Erot.

KE1=KE2+Erot

1/2mvi^2=1/2mvf^2+1/2(2/5m)(vf^2/r^2)

After mathing, I solved that the final velocity is 1.30559 m/s, but that isn't right.

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#### robphy

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You can apply the conservation of total energy if you can properly account for the changes (transfers) of energy during the process from start to finish.

#### kingsmaug

*facepalm* Energy loss due to friction.

So,

KEi = KEf + Ertof - Efric

1/2mvi^2=1/2mvf^2+1/2(2/5m)(vf^2/r^2) - uk(N)(d)

Which when solved gives me vf=1.59 m/s, which is also wrong.

#### robphy

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KEi = KEf + Ertof - Efric

1/2mvi^2=1/2mvf^2+1/2(2/5m)(vf^2/r^2) - uk(N)(d)
Is your rotational kinetic energy correct? (do the units makes sense?)

#### kingsmaug

Units, yes. But I just realized that I left out the r^2 term from I.

So the rotational energy is 1/2*I*w^2, and w=v/r

That turns to 1/2*I*(v/r)^2

But I suppose it doesn't entirely matter anymore. I've run out of attempts on the problem.

#### robphy

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But I suppose it doesn't entirely matter anymore. I've run out of attempts on the problem.
Possibly,... if you never have to solve a problem like that in the future.

#### ehild

Homework Helper
The friction decelerates the linear velocity according to v=Vo-at. If you know the value of a, and also the time of skidding, t, why do you not use the formula v=Vo -at to get the final linear velocity?

Last edited:

#### haruspex

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2018 Award
The easiest way to solve this is by considering angular momentum. The trick is to choose a reference axis such that you don't care about the frictional force.

#### physics09

The magnitude of the final velocity can be solved for by using v=vi-ugt, where t=(2vi)/(7ug).
Substitute t into the formula and solve for v.
Remember, vi is given to you in the problem.

legend:
u=mu
g=gravity
t=time
vi=initial velocity
v=what you're solving for

#### haruspex

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2018 Award
The magnitude of the final velocity can be solved for by using v=vi-ugt, where t=(2vi)/(7ug).
Substitute t into the formula and solve for v.
Remember, vi is given to you in the problem.

legend:
u=mu
g=gravity
t=time
vi=initial velocity
v=what you're solving for
As I posted, there's no need to find the time if all you care about is final velocity. Just use conservation of angular momentum.

#### dean barry

if you ignore frictional losses, then if you assume the ball is a solid homogenous sphere, the final ratio of linear KE to rotational KE will be:
5/7 : 2/7
This totaled will match the original linear KE

#### dean barry

if you include friction converted to heat:
As linear deceleration and rotational acceleration are constant, then the driving forces are also deemed to be constant, if you work these out from the figures, the rest of the data should fall into place.
Use time elapsed as a guide, calculate the point in time where diminishing linear velocity and increasing rotational surface velocity are equal.
(synchronous speed)

#### haruspex

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2018 Award
if you ignore frictional losses,
Why would you do that? It will clearly produce a wrong answer. In particular, it will violate other conservation laws.

#### dean barry

im going from the given numbers
the time given (0.824 s) and the derived distance (5.79 m) are correct if you use the method i outlined (post 12)
I did it by trial and error based on elapsed time and you can then figure the final velocity using:
elapsed time (0.824 s)
initial velocity (8.2 m/s)
and linear deceleration rate ( - 2.84 m/s^2 )
by using the appropriate newtons equation of motion
with the final velocity you can calculate the linear KE and then use the 5/7 : 2/7 rule to find the rotational KE

#### haruspex

Homework Helper
Gold Member
2018 Award
im going from the given numbers
the time given (0.824 s) and the derived distance (5.79 m) are correct if you use the method i outlined (post 12)
I did it by trial and error based on elapsed time and you can then figure the final velocity using:
elapsed time (0.824 s)
initial velocity (8.2 m/s)
and linear deceleration rate ( - 2.84 m/s^2 )
by using the appropriate newtons equation of motion
with the final velocity you can calculate the linear KE and then use the 5/7 : 2/7 rule to find the rotational KE
Or just use conservation of angular momentum about a point on the ground:
$m v_i R = m v_f R + I v_f / R = m v_f R(1+\frac 25)$
Note that the mass, radius and coefficient of friction are all irrelevant.